Prove that $p$ is a limit point of A.

Question

Let $p$ be a limit point of $\bar{A}$ where $\bar{A}$ is the set of all limit points of $A$. Show that $p$ is also a limit point of $A$.

According to the definition of a limit point I know that I have to show that $B(p; \delta ) \cap A \neq \emptyset$ for all $\delta > 0$. I know that $B(p; \delta ) \cap \bar{A} \neq \emptyset$. Because this set is not empty, I know I can find a point $q$ for every $\delta$ in this set, so also for $\frac{\delta}{2}$. I know I have to look at the set

$B(q; \frac{\delta}{2}) \cap A$, but I am not sure how I can use this to prove that $p$ is a limit point of $A$.

Answer 1

Let $\varepsilon>0$; you want to prove that there is a $a\in A\setminus\{p\}$ such that $d(a,p)<\varepsilon$. Since $p\in\overline A$, there is a $a'\in\overline A\setminus\{p\}$ such that $d(a',p)<\frac\varepsilon2$. And, since $a'\in\overline A$, there is a $a\in A$ such that $d(a',a)<\frac\varepsilon2$. Clearly, you can take $a\neq p$. But then$$d(a,p)\leqslant d(a,a')+d(a',p)<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$

Answer 2

Let $x\in B(p,\delta)\cap\bar A$, write $r=d(x,p)$there exists $y\in B(x,{{\delta-r}\over 2})\cap A$, we have $d(p,y)\leq d(p,x)+d(x,y)\leq r+{{\delta-r}\over 2}$

$\delta-(r+{{\delta -r}\over 2})={{\delta-r}\over 2}$, implies that $y\in B(p,\delta)$.